Question

Determine the zero(s) of the quadratic function y = 2x^2 = 9x + 4.

Answer 1

To find the zero(s) of the quadratic function \( y = 2x^2 - 9x - 4 \), we can set the function equal to zero:

\[
2x^2 - 9x - 4 = 0
\]

We will use the quadratic formula, which is given by:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

where \( a = 2 \), \( b = -9 \), and \( c = -4 \).

First, we need to calculate the discriminant \( b^2 - 4ac \):

\[
b^2 = (-9)^2 = 81
\]
\[
4ac = 4 \cdot 2 \cdot (-4) = -32
\]

Now, we compute the discriminant:

\[
b^2 - 4ac = 81 - (-32) = 81 + 32 = 113
\]

Since the discriminant is positive, we can expect two real and distinct solutions. Now let's substitute \( a \), \( b \), and the discriminant into the quadratic formula:

\[
x = \frac{-(-9) \pm \sqrt{113}}{2 \cdot 2}
\]

This simplifies to:

\[
x = \frac{9 \pm \sqrt{113}}{4}
\]

Thus, the zeros of the quadratic function are:

\[
x_1 = \frac{9 + \sqrt{113}}{4}
\]

and

\[
x_2 = \frac{9 - \sqrt{113}}{4}
\]

So, the zeros of the function \( y = 2x^2 - 9x - 4 \) are \( x_1 = \frac{9 + \sqrt{113}}{4} \) and \( x_2 = \frac{9 - \sqrt{113}}{4} \).